如何在python中使用正则表达式删除字母并提取数字？

Question

How to remove alphabets and extract numbers using regex in python?

import re
l=["098765432123 M","123456789012"]
s = re.findall(r"(?<!\d)\d{12}", l)
print(s)

Expected Output:

123456789012

Answer 1

If all you want is to have filtered list, consisting elements with pure digits, use filter with str.isdigit :

list(filter(str.isdigit, l))

Or as @tobias_k suggested, list comprehension is always your friend:

[s for s in l if s.isdigit()]

Output:

['123456789012']

Answer 2

To keep only digits you can do re.findall('\\d',s) , but you'll get a list:

s = re.findall('\d', "098765432123 M")
print(s)
> ['0', '9', '8', '7', '6', '5', '4', '3', '2', '1', '2', '3']

Answer 3

I would suggest to use a negative lookahead assertion, if as stated you want to use regex only.

l=["098765432123 M","123456789012"]
res=[]
for a in l:
    s = re.search(r"(?<!\d)\d{12}(?! [a-zA-Z])", a)
    if s is not None:
        res.append(s.group(0))

The result would then be:

['123456789012']

Answer 4

So to be clear, you want to ignore the whole string if there is a alphabetic character in it? Or do you still want to extract the numbers of a string with both numbers and alphabetic characters in it?

If you want to find all numbers, and always find the longest number use this:

regex = r"\d+"
matches = re.finditer(regex, test_str, re.MULTILINE)

\\d will search for digits, + will find one or more of the defined characters, and will always find the longest consecutive line of these characters.

If you only want to find strings without alphabets:

import re
regex = r"[a-zA-Z]"
test_str = ("098765432123 M", "123456789012")
for x in test_str:
    if not re.search(regex, x):
        print(x)